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b. What is the initial velocity of the bullet? c. What is initial kinetic energy of the bullet-block system before collision? 2. A 15-g bullet is shot vertically into an 2-kg block. The block lifts upward 8.0 mm (see the figure). The bullet penetrates the block and comes to rest in it in a time interval of 0.0010 s. Assume the force on the bullet is constant dining penetration and that air resistance is negligible. a. What is the velocity of the bullet-plus-block system right after the collision?

Explanation / Answer

A) (1/2)*(m+M)*v^2 = (m+M)*h*g...

v = sqrt(2*g*h) = sqrt(2*9.8*0.008) = 0.3959 m/sec...

B) m*u = (m+M)*v = (0.015+2)*0.3959 =0.7977

initial velocity is u = 0.7977/m = 0.7977/0.015 = 53.18 m/sec...

C)Kinitial = (0.5*m*u^2)+0 = 21.21 J

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